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raketka [301]
3 years ago
8

Please help I need a 100

Chemistry
2 answers:
kolbaska11 [484]3 years ago
8 0

Answer:C

Explanation:

The molcules of a liquid have middle space between them,but lesser than gas molecules.

IT is due to liquids have stroger van der waal forces then gases.

Thier kinetic energy also depends on molecular motion,which is lesser than gas molecules and hence the liquids have lesser kinetic energy than gases and plasma and slightly more than solids.

Solid molecules have closest packing hence strogest force of attraction,which will decrease the kinetic energy that is K.E near to zero.

musickatia [10]3 years ago
7 0
C has to be the correct answer because solids have the slowest moving molecules, and gas molecules move faster than liquid molecules.
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Answer:

344.21 g/mol

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

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Explanation:

<u>Step 1: Define</u>

Cr₂(SO₃)₃

<u>Step 2: Identify</u>

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<u>Step 3: Find</u>

Molar Mass of Cr₂(SO₃)₃ - 2(52.00) + 3(32.07) + 9(16.00) = 344.21 g/mol

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3 years ago
Will P and O2 share or transfer electrons?<br> P + O2 →P4,O10
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They will share electrons .

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Which element has 32 protons in its nucleus?
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Calculate the total amount of energy required to change 10.0 g of water from 35.0 degrees Celsius to 110. degrees Celsius.
Makovka662 [10]

Answer:

The total amount of energy required is 25,515.2 J.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When a system absorbs (or gives up) a certain amount of heat, it can happen that:

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To calculate the latent heat the formula is used:

Q = m. L

Where

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To calculate sensible heat the following formula is used:

Q = m. c. ΔT

where:

  • Q: amount of sensible heat  
  • m: body mass
  • c: specific heat of the substance
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In this case, you have in the first place a heat to raise the temp of the water from 35.0 C to 100 C, where the specific heat value for water is  4.184 \frac{J}{g*C}:

q1 = m*c*(Tfinal-Tinitial)

q1 = 10.0 g *(4.184 \frac{J}{g*C})* (100 - 35.0 C) = 2719.6 J

Now you have the heat to vaporize the water, where the heat of vaporization is 2259.36 \frac{J}{g}:

q2 = m*(heat of vaporization)

q2 = 10.0 g*(2259.36 \frac{J}{g}) = 22593.6 J

Finally, you have the heat to raise temp of steam to 110 C, where the specific heat value for steam is  2.02 \frac{J}{g*C} :

q3 = m*c*(Tfinal-Tinitial)

q3 = 10.0 g*(2.02 \frac{J}{g*C})*(110-100 C) = 202 J

The total amount of energy can be calculated as:

Q= q1 + q2 + q3

Q= 2719.6 J + 22593.6 J + 202 J

Q=25,515.2 J

<u><em>The total amount of energy required is 25,515.2 J.</em></u>

5 0
3 years ago
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